In the adjoining figure the impedance (in ohm) of the circuit will be

${\mathrm{i}}_{\mathrm{L}}=\frac{90}{30}=3\mathrm{A}, {\mathrm{i}}_{\mathrm{C}}=\frac{90}{20}=4.5\mathrm{A}$

Net current through circuit  $\mathrm{i}={\mathrm{i}}_{\mathrm{C}}-{\mathrm{i}}_{\mathrm{L}}=1.5\text{\hspace{0.33em}}\mathrm{A}$

$\therefore \mathrm{Z}=\frac{\mathrm{V}}{\mathrm{i}}=\frac{90}{1.5}=60\mathrm{\Omega }$